Question

A particle of mass $m$ just completes the vertical circular motion. What will be the difference in tension at the lowest and highest point?

• A. $2mg$
• B. $4mg$
• C. $8mg$
• D. $6mg$

Answer 1

The correct option is D $6mg$


Let a small body of mass $m$ attached to a string revolve in a vertical circle of radius $r$.
Let $v_2$ be the speed of the body and $T_2$ be the tension in the string at the lowest point $B$. So at the lowest point
$T_2=\frac{m{v}_2^2}{r}+mg$ .......(1)
Total energy at bottom = $PE+KE$
$=0+\frac{1}{2}m{v}_2^2$ $=\frac{1}{2}m{v}_2^2$ .........(2)

Let $v_1$ be the speed and $T_1$ be the tension in the string at highest point $A$.
So, at highest point $A$, $T_1=\frac{m{v}_1^2}{r}-mg$ .......(3)
Total energy at $A$ = $PE+KE$
$=2mgr+\frac{1}{2}m{v}_1^2$ ......(4)

From equations (1) and (3)
$T_2-T_1=\frac{m{v}_2^2}{r}+mg-(\frac{m{v}_1^2}{r}-mg)$
$=\frac{m}{r}(v_2^2-v_1^2)+2mg$ ........(5)

By law of conservation of energy
Total Energy at A = Total energy at B.
From equations (2) and (4)
$\frac{1}{2}m{v}_2^2=\frac{1}{2}m{v}_1^2+2mgr$
So $v_2^2-v_1^2=4rg$ .........(6)
Putting this value in equation (5)
$T_2-T_1=\frac{m}{r}\left(4rg\right)+2mg$
= $4mg+2mg$
$\therefore T_2-T_1=6mg$